In a meter bridge experiment, S is a standard resistance, R is a resistance wire. It is found that b

Q: Why is the length of the right side of the wire taken as $100 - l$?

A standard meter bridge uses a wire that is exactly 1 meter, or 100 cm, long. If the balance point is at a distance $l$ from the left end, the remaining length of the wire must be $100 - l$.

Q: Why does halving the diameter increase the resistance so much?

Resistance is inversely proportional to the cross-sectional area. Since area depends on the square of the diameter ($A = \pi d^2 / 4$), halving the diameter reduces the area by a factor of 4. This causes the resistance to increase by a factor of 4, before accounting for the change in length.

Q: Can we use radius instead of diameter in the resistance formula?

Absolutely! The radius is simply half the diameter. If the diameter is halved, the radius is also halved. Since area is $\pi r^2$, halving the radius reduces the area by a factor of $2^2 = 4$. The result remains exactly the same.

Q: What happens to the standard resistance $S$ when $R$ is replaced?

The standard resistance $S$ remains completely unchanged. It is a fixed resistor in the right gap. Only the wire in the left gap was replaced, which is why we only calculated a new value for $R'$ and kept $S$ the same in our final equation.

Q: Does the material of the meter bridge wire affect the balance point?

No, as long as the wire is uniform. The balance condition $\frac{R}{S} = \frac{l}{100-l}$ relies on the ratio of the resistances of the two segments of the wire. Since both segments are made of the same material and have the same cross-sectional area, their resistivity and area cancel out, leaving only the ratio of their lengths.

Question

In a meter bridge experiment, $S$ is a standard resistance, $R$ is a resistance wire. It is found that balancing length is $l = 25	ext{ cm}$. If $R$ is replaced by a wire of half length and half diameter that of $R$ of same material, then the balancing distance $l'$ (in cm) will now be ..........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a physics lab, looking at a classic meter bridge setup. A meter bridge is a practical application of the Wheatstone bridge principle, used to find unknown resistances. In our left gap, we have a resistance wire $R$, and in the right gap, a standard resistance $S$.

The problem states that initially, the bridge is balanced at a length $l = 25	ext{ cm}$ from the left end. The fundamental principle of a balanced meter bridge tells us that the ratio of the resistances in the gaps is equal to the ratio of the lengths of the wire segments they balance against.

Mathematically, this is written as:

$$ \frac{R}{l} = \frac{S}{100 - l} $$

Substituting our known value of $l = 25	ext{ cm}$, we get:

$$ \frac{R}{25} = \frac{S}{100 - 25} $$

$$ \frac{R}{25} = \frac{S}{75} $$

Simplifying this fraction, we find a direct relationship between our two resistors:

$$ R = \frac{S}{3} \implies S = 3R $$

### The Master Equation: Changing the Wire

Now comes the twist. The original wire $R$ is removed and replaced by a new wire of the **same material**, but with **half the length** and **half the diameter**. Let's call this new resistance $R'$. To find $R'$, we need to recall the formula that connects resistance to the physical dimensions of a wire:

$$ R = ho \frac{L}{A} $$

Here, $ho$ is the resistivity (which stays constant since the material is the same), $L$ is the length, and $A$ is the cross-sectional area. Since the wire is cylindrical, its area is given by $A = \frac{\pi d^2}{4}$, where $d$ is the diameter. Substituting this into our resistance formula gives:

$$ R = \frac{4ho L}{\pi d^2} $$

Now, let's carefully plug in the new dimensions for our replacement wire. The new length is $L' = \frac{L}{2}$ and the new diameter is $d' = \frac{d}{2}$.

$$ R' = \frac{4ho (L/2)}{\pi (d/2)^2} $$

When we square the denominator, $(d/2)^2$ becomes $d^2/4$. The $4$ flips up to the numerator, multiplying with the existing terms:

$$ R' = \frac{2ho L}{\pi d^2 / 4} = 8 \frac{ho L}{\pi d^2} $$

Notice how this relates to our original resistance $R$. We can factor out a $2$ to see the original expression:

$$ R' = 2 \left( \frac{4ho L}{\pi d^2} ight) = 2R $$

Fascinating! By halving both the length and the diameter, the resistance actually **doubles**. This happens because the resistance is inversely proportional to the *square* of the diameter, which dominates the linear decrease in length.

### Final Calculation: The New Balance Point

With our new resistance $R' = 2R$ in the left gap, the balance point will inevitably shift. Let's call the new balancing length $l'$. We apply the Wheatstone bridge principle one more time:

$$ \frac{R'}{l'} = \frac{S}{100 - l'} $$

Substitute $R' = 2R$ and our previously found relationship $S = 3R$:

$$ \frac{2R}{l'} = \frac{3R}{100 - l'} $$

The beauty of this setup is that the unknown resistance $R$ cancels out perfectly from both sides, leaving us with a simple linear equation:

$$ \frac{2}{l'} = \frac{3}{100 - l'} $$

Cross-multiplying yields:

$$ 2(100 - l') = 3l' $$

$$ 200 - 2l' = 3l' $$

Bringing the $l'$ terms to one side:

$$ 5l' = 200 $$

$$ l' = 40	ext{ cm} $$

And there we have it! The new balancing distance is exactly **40 cm**.

Enter Numerical Value:

Visualized Solution

$l = 25 cm$

$\frac{R}{l} = \frac{S}{100 - l}$

$\frac{R}{25} = \frac{S}{75}$

$S = 3 R$

$L^{'} = \frac{L}{2}, d^{'} = \frac{d}{2}$

$R = ρ \frac{L}{A}$

$R^{'} = \frac{4 ρ ( L /2 )}{π ( d /2 ) ^{2}}$

$R^{'} = 2 R$

$l^{'}$

$\frac{2 R}{l ^{'}} = \frac{3 R}{100 - l ^{'}}$

$2 (100 - l^{'}) = 3 l^{'}$

$l^{'} = 40 cm$

\text{Conclusion}

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l=25 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } lR​=100−lS​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 25R​=75S​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=3R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L′=2L​,d′=2d​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=ρAL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R′=π(d/2)24ρ(L/2)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R′=2R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l′2R​=100−l′3R​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2(100−l′)=3l′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l′=40 cm